Question

How do you find the derivative of `f(x)=log_2(x^2/(x-1))`?

Answer 1

`f'(x)=(x-2)/(x(x-1)(ln(2)))`

Explanation:

We need to know the logarithm rules:

• `log_a(b/c)=log_a(b)-log_a(c)`
• `log_a(b^c)=clog_a(b)`
• `log_a(b)=log_c(b)/log_c(a)=ln(b)/ln(a)`

We also have to know that:

• `d/dx(ln(x))=1/x`
• `d/dxln(f(x))=1/f(x)*f'(x)`

Then:

`f(x)=log_2(x^2)-log_2(x-1)`

`f(x)=ln(x^2)/ln(2)-ln(x-1)/ln(2)`

`f(x)=1/ln(2)(2ln(x)-ln(x-1))`

The derivative is then:

`f'(x)=1/ln(2)(2(1/x)-1/(x-1)(x-1)')`

`f'(x)=1/ln(2)(2/x-1/(x-1))`

Which can be simplified:

`f'(x)=1/ln(2)((2(x-1)-x)/(x(x-1)))`

`f'(x)=(x-2)/(x(x-1)(ln(2)))`